MHB Noetherian Rings: Does R Need to Be Finitely Generated? - Peter

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In the discussion about Noetherian rings, the definition states that a commutative ring R is Noetherian if every ideal of R is finitely generated. The question raised is whether R itself must be finitely generated since it is an ideal of itself. This leads to confusion in the context of fields, as every field is considered Noetherian, yet it seems incorrect to assert that every field is finitely generated. The clarification provided indicates that while R can be viewed as an R-vector space with a basis of {1}, this does not imply that R is finitely generated in the traditional sense. Thus, the nuances of Noetherian properties in relation to fields and their ideals need careful consideration.
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Dummit and Foote in Chapter 9 - Polynomial Rings define Noetherian rings as follows:

Definition. A commutative ring R with 1 is called Noetherian if every ideal of R is finitely generated.

Question: Does this mean that R itself must be finitely generated since R is an ideal of R?

This question is important in the context of fields since D&F go on to say that every field is Noetherian. In the case of fields the only ideals are the trivial ideal {0} and the field itself. But this would mean every field is finitely generated which does not seem to be corect.

Can anyone clarify these issues for me?

Peter

[This has also been posted on MHF]
 
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Peter said:
Question: Does this mean that R itself must be finitely generated since R is an ideal of R?

Take into account that for every commutative and unitary ring $R$ (noetherian or not), we have $R=R\cdot 1=(1)$.

But this would mean every field is finitely generated which does not seem to be corect.

Why? In such a case, $R$ is an $R$-vector space and a basis is $B=\{1\}$.
 
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